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Applications of Integrations #11 [Solved!]

My question

I haven't had calculus in many years but I like to work problems every now and then to stay sharp. I have a problem which is almost identical to your item 11 Arc Length of a Curve using integration. I am following the logic all the way through example 1 down to the answer. It states Y = 1.35 sin 0.589x so dy/dx = 0.795 cos 0.589x. How did the jump happen between these equations? I don't follow how 1.35 sin becomes 0.795 cos.

Relevant page

11. Arc Length of a Curve using Integration

What I've done so far

Whacking my head against a wall :)

I understand the logic of a and b and the move to the integral but cannot understand where 0.795 cos came from.

X

I haven't had calculus in many years but I like to work problems every now and then to stay sharp.  I have a problem which is almost identical to your item 11 Arc Length of a Curve using integration.  I am following the logic all the way through example 1 down to the answer.  It states Y = 1.35 sin 0.589x so dy/dx = 0.795 cos 0.589x.  How did the jump happen between these equations? I don't follow how 1.35 sin becomes 0.795 cos.
Relevant page

<a href="/applications-integration/11-arc-length-curve.php">11. Arc Length of a Curve using Integration</a>

What I've done so far

Whacking my head against a wall :)

I understand the logic of a and b and the move to the integral but cannot understand where 0.795 cos came from.

Re: Applications of Integrations #11

@Kabookiep: The background can be found here: Differentiation of Sin, Cos, Tan.

It's actually this case:

`dy/dx=dy/(du) (du)/dx`

For that example, we have `y=sin u` where:

`u=0.598x` so `(du)/dx=0.598`

and

`dy/(du) = 1.35 cos u = 1.35 cos 0.598x`

Thus

`dy/dx=[1.35cos(0.598x)] xx (0.598)`

`= 0.795 cos 0.598x`

X

@Kabookiep: The background can be found here: <a href="https://www.intmath.com/differentiation-transcendental/1-derivative-sine-cosine-tangent.php">Differentiation of Sin, Cos, Tan</a><a></a>.

It's actually this case:

`dy/dx=dy/(du) (du)/dx` 

For that example, we have `y=sin u` where:

`u=0.598x` so `(du)/dx=0.598`

and

`dy/(du) = 1.35 cos u = 1.35 cos 0.598x`

Thus

`dy/dx=[1.35cos(0.598x)] xx (0.598)`

`= 0.795 cos 0.598x`

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